1:25 AM
hi

8 hours later…
9:40 AM
Hi @hellofriends

2 hours later…
11:38 AM
If $f:D\to\Bbb C$ is a locally univalent map on a simply connected domain $D\subset\Bbb C$, then $f$ is globally univalent map?
11:54 AM
@onepotatotwopotato won't $\exp:\mathbb{C}\to\mathbb{C}$ be a counter-example?
$\exp(z) = \exp(w)$ iff $z-w\in 2\pi\cdot i\cdot \mathbb{Z}$ so lets choose for $z\in\mathbb{C}$ a ball $B(z, \pi)$
12:49 PM
Only two online, that's rare
1:38 PM
I am asked to solve an odd exercise I think (I don't see the purpose in this exercise actually, since transforms just make it harder I think). So the exercise reads;
> if $X\mid \Sigma^2=\lambda\in N(0,1/\lambda)$ with $\Sigma ^2\in \Gamma \left(\frac{n}{2}{,}\frac{2}{n}\right)$, show that $X\in t(n)$ using transforms.
Here's my attempt; denote by $\varphi_X$ the cf of $X$, then $$\varphi_X(t)=Ee^{itX}=E(E(e^{itX}\mid\Sigma^2))=Eh(\Sigma^2),$$where $h(\lambda)=\varphi_{X\mid \Sigma^2=\lambda}(t)=e^{-t^2/(2\lambda^2)}$, since recall the cf of $N(0,\sigma^2)$ is just $e^{-t^2\sigma^2/2}$. Now, $$\varphi_X(t)=Ee^{-t^2/(2\Sigma^2)}=\ldots,$$and I don't know how to proceed further.
The thing is, I don't want to take this route actually, since the cf of $t(n)$ is very intricate, but I guess this is the way to go, by comparing cfs, or?
EDIT: I notice a typo; $h(\lambda)$ should be just $e^{-t^2/(2\lambda)}$.
@SoumikMukherjee chat is more empty than my brain lately
2:25 PM
0

The last claim is not necessarily true. If you treat a metric space as a topological space, you can define a U and a W based on epsilon-balls that both contain p, but have a different point as a center.

is this computer generated? Or am I just not smart enough to understand this answer
2:40 PM
the question itself doesn't matter for the site at large of course, its a duplicate upon duplicate of questions about local connectedness
3:38 PM
@SineoftheTime ☠️
3:51 PM
I am reading a proof and the following claim is made: Because $X$ is a compact metric space, the closed ball $B$ is contained in only finitely closed balls of $X$.
I tried verifying this but I was unsuccessful...I could use a hint or something.
@user193319 that's false though?
Seems false or stated improperly. If $B = B(x,r)$, then $B$ is contained in $B(x,\delta)$ for any $\delta > r$, of which there are infinitely many of them.
Yeah, that's what I was thinking...
the closed ball of $\mathbb{R}$, $[0, 1]$, is contained in infinitely many closed balls
any interval $[a, b]$ with $a \leq 0$ and $1\leq b$ works
The claim comes from here: citeseerx.ist.psu.edu/…
what's the language of the assertion exactly?
3:54 PM
See lemma 3.7 (I guess I used slightly different notation).
@user193319 which page
Page 6
I'll need some time to digest assumptions here, since I'm not experienced with the topic
I can establish the existence of a region for the group element and a point, so that part you can ignore. I just don't understand the argument that we can from there construct a maximal region.
@Jakobian Okie dokie. I appreciate any help. I'll work through the rest of the paper in the meantime...if something comes to me, I'll let you know.
$\Gamma$ is some group acting on $X$?
3:57 PM
@Jakobian Yes, but not necessarily by isometries but "local similiarities"
oh okay so $X$ is an ultrametric space
Yes...maybe the assertion is true for such spaces? I was thinking he was making a general assertion about compact metric spaces.
@user193319 ultrametric spaces have the property that intersecting balls are contained in each other
this should mean then, that the balls that contain $x$ form a chain
Yes, for open and closed balls.
Ah, yes!
There is an upper bound on the radii because the metric space is compact and hence has finite diameter, right?
If there was an infinite amount of closed balls $B_n = \overline{B}(x, r_n)$ with $r_n$ increasing, then since $B_{n+1}\setminus B_n$ is clopen, we would have an infinite open cover without a finite subcover
4:08 PM
Ah, yes...nice. Ultrametric spaces are weird.
the chain argument works as well I guess, although it proves less
this is confusing to me as well, since Cantor set is an ultrametric space, and I'm not very comfortable with there being a finite amount of closed/open balls
also above I don't really have the comfort of letting $r_n$ be increasing, but the argument works
oh okay I see, this argument doesn't actually work
I was thinking about $\{0\}\cup \{1/n : n\in\mathbb{N}\}$ but thats not an ultrametric space, athough ultrametrizable
If there is an infinite amount of distinct balls $B_n$ containing $x$, we can take a monotone subsequence of the radiuses and then proceed like how I wanted to above
so this does use that 1) they are clopen, 2) they form a chain
no this still doesn't work if the radiuses are strictly decreasing I suppose since you'd have $B_{n-1}\setminus B_n$ but you'd have to account for $\bigcap_n B_n$ and I don't think thats necessarily clopen
oh no it is clopen because those are closed balls so this is also a closed ball, okay good
and this is also where we need that they are all contained in a ball
@user193319 Do you understand me? If not, I can try to repeat the argument more clearly
5:10 PM
@Jakobian Hmm. Yeah, I wouldn't mind seeing the argument written out again. Thank you!
5:21 PM
@user193319 Consider all the closed balls that contain $B$. All of them have $x$ as center and form a chain. We can pick a set $A\subseteq (0, \infty)$ of radiuses such that each ball containing $B$ is of the form $\overline{B}(x, r)$ for some $r\in A$, and for two distinct $s\neq r$ from $A$, $B(x, r)\neq B(s, r)$. Then $A\subseteq [a, b]$ for some $0 < a < b$ since $X$ is compact and all of the balls $\overline{B}(x, r)$ for $r\in A$ must contain $B$
Take a monotone sequence $a_n\in A$ with $a_n\to a\in (0, \infty)$. If $a_n$ is strictly increasing, then $\overline{B}(x, a_1)$ and $\overline{B}(x, a_{n+1})\setminus \overline{B}(x, a_n)$ and $X\setminus B(x, a)$ form an infinite partition of $X$, so it has no finite subcover, this is since all of the balls in an ultrametric space are open
If $a_n$ is decreasing, take $X\setminus \overline{B}(x, a_1)$ and $\overline{B}(x, a_n)\setminus \overline{B}(x, a_{n+1})$ and $\overline{B}(x, a)$. Then this is again an infinite partition of $X$ by open sets without a finite subcover
7 mins ago, by Jakobian
@user193319 Consider all the closed balls that contain $B$. All of them have $x$ as center and form a chain. We can pick a set $A\subseteq (0, \infty)$ of radiuses such that each ball containing $B$ is of the form $\overline{B}(x, r)$ for some $r\in A$, and for two distinct $s\neq r$ from $A$, $B(x, r)\neq B(s, r)$. Then $A\subseteq [a, b]$ for some $0 < a < b$ since $X$ is compact and all of the balls $\overline{B}(x, r)$ for $r\in A$ must contain $B$
in here the two balls $B(x, r)\neq B(s, r)$ should be $\overline{B}(x, r)\neq \overline{B}(x, s)$
everywhere else when I write without overline I mean an open ball
5 mins ago, by Jakobian
Take a monotone sequence $a_n\in A$ with $a_n\to a\in (0, \infty)$. If $a_n$ is strictly increasing, then $\overline{B}(x, a_1)$ and $\overline{B}(x, a_{n+1})\setminus \overline{B}(x, a_n)$ and $X\setminus B(x, a)$ form an infinite partition of $X$, so it has no finite subcover, this is since all of the balls in an ultrametric space are open
here monotone means strictly monotone - I am assuming $A$ is infinite
since both cases lead to a contradiction, $A$ needs to be finite, so in a compact ultrametric space, for any closed ball $B$, there exists only finite amount of closed balls containing $B$
Ah, very nice! This was a bit more complicated than I expected!
@Jakobian Thanks!
5:54 PM
@Jakobian I actually have a question about definition 3.3 and proposition 3.5 that was bugging me before. For $g$ to be locally determined by $\text{Sim}_{X}$, does definition 3.3 have to hold for some closed ball or every closed ball? It's not really clear to me.
6:05 PM
@user193319 you mean $B\subseteq X$?
and if it has to go over all balls $B$?
Yes, the closed ball B referenced in definition 3.3
how I understand it is that they are applying the definition 3.3 to the closed ball $X$
note that this is a closed ball since $X$ is compact
Ah, that makes more sense!

2 hours later…
8:16 PM
Is there a good intuition for why a primal constrained minimization problem is equivalent to the max over the dual variables of the min over the primal variables of the Lagrangian?
8:49 PM
I am studying sums of i.i.d random variables where the number of terms is random, i.e. we have $S_n=X_1+\ldots+X_n$ and then we put $S_N$ where $N$ is a nonnegative integer-valued random variable. I am stuck at a derivation of the generating function for $S_N$. Recall, the generating function of a r.v. $X$ is $g_X(t)=Et^X$. Now, $$g_{S_N}(t)=Et^{S_N}=\sum_{n=0}^\infty E(t^{S_N}\mid N=n)\cdot P(N=n)=\ldots.$$I don't understand the second equality here. Which definition of expectation is this?
Above, $X_1,X_2,\ldots$ are also assumed to be nonnegative, integer-valued.
Also, $N$ is independent of $X_1,X_2,\ldots$.
@psie Conditional expectation?
I don't know actually, just guessing
9:31 PM
@SineoftheTime almost ;) it's the law of iterated expectation I think, i.e. $$E(t^{S_N}) = E(E(t^{S_N} \mid N)).$$
yes
9:57 PM
@Thorgott what's the most intuitive proof you've seen of the fact that if $X, Y$ are $T_4$, $A\subseteq X$ is closed and $f:A\to Y$ is continuous, then $Y\cup_f X$ is $T_4$? Dugundji does it by playing with open sets, doesn't seem too intuitive. I've also seen some proof by proving that Tietze extension theorem holds for $Y\cup_f X$
I've tried proving it myself by using Tietze and Urysohn lemma, but it seems pretty complicated and I didn't manage to finish the proof
My attempt was like this, let $F_1, F_2\subseteq Y\cup_f X$ be disjoint closed sets, let $F_i' = q^{-1}[F_i]$ where $q$ is the quotient map and $F_i' = A_i\cup B_i$ where $A_i\subseteq Y$ and $B_i\subseteq X$ are closed where $f^{-1}[A_i] = A\cap B_i$ since $F_i'$ are saturated. Find $h:Y\to [0, 1]$ separating $A_1$ and $A_2$, $g = h\circ f$ and $\tilde{g}$ an extension of $g$ to $X$.
Last time I had problem with obtaining disjoint neighbourhoods of $B_1, B_2$ so that everything is saturated and disjoint
11:01 PM
Actually after going through some details, I think the proof by showing that every continuous $g:F\to\mathbb{R}$ where $F\subseteq Y\cup_f X$ is closed can be extended isn't that bad
11:30 PM
I just noticed that every involutive group is abelian. $xy = (yx)(yx)xy = yxy(xx)y = yx(yy) = yx$.
@DannyuNDos $(xy)^2 = xy \implies xy = yx$
Huh? The notion of involutions should give $(xy)^2 = e$.
11:45 PM
Yeah sorry. $(xy)^2 = e \implies x(xy)^2y = xy \implies yx = xy$
Yeah, nice.